Question

(a) find an equation of the tangent line to the graph of \(f\) at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ g(x)=x+e^{x} $$ $$ (0,1) $$

Short answer

The equation of the tangent line to the graph of the function \(g(x) = x + e^{x}\) at the point (0,1) is \(y = 2x + 1\).

Step-by-step solution

Calculate the derivative of the function

The derivative of the function \(g(x) = x + e^{x}\) can be found using the power rule for differentiation, as well as the fact that the derivative of \(e^{x}\) is \(e^{x}\) itself. The derivative of \(g(x)\) is therefore \(g'(x) = 1 + e^{x}\).

Find the slope of the tangent line at the given point

Substituting the x-coordinate of the given point (0,1) into the derivative function provides the slope of the tangent line at that point. So, \(g'(0) = 1 + e^{0} = 1 + 1 = 2\). Thus, the slope of the tangent line at the point (0,1) is 2.

Derive the equation of the tangent line at the given point

With the slope (\(m = 2\)) of the tangent line and a point on it (\((x_{1}, y_{1}) = (0, 1)\)), the equation of the tangent line can be found using the formula \(y - y_{1} = m(x - x_{1})\). Plugging the given values into this formula provides the equation of the tangent line: \(y - 1 = 2(x - 0)\), or, in the simplified form, \(y = 2x + 1\).